Solve for $x$ : $9\sqrt{x} + 1 = 2\sqrt{x} + 3$
Solution: Subtract $2\sqrt{x}$ from both sides: $(9\sqrt{x} + 1) - 2\sqrt{x} = (2\sqrt{x} + 3) - 2\sqrt{x}$ $7\sqrt{x} + 1 = 3$ Subtract $1$ from both sides: $(7\sqrt{x} + 1) - 1 = 3 - 1$ $7\sqrt{x} = 2$ Divide both sides by $7$ $\frac{7\sqrt{x}}{7} = \frac{2}{7}$ Simplify. $\sqrt{x} = \dfrac{2}{7}$ Square both sides. $\sqrt{x} \cdot \sqrt{x} = \dfrac{2}{7} \cdot \dfrac{2}{7}$ $x = \dfrac{4}{49}$